Remark 1 
Consider a condition \(\mathfrak {M}\subset \mathrm {D}( W ^{\ast }),\)  in this case the operator \(\mathcal {H}:=\mathfrak {Re }\,W\)  is defined on \(\mathfrak {M},\)  the fact is that \(\tilde{\mathcal {H}}\)  is selfadjoint, bounded from bellow (see Lemma 3 [1]}). Hence a corresponding sesquilinear form (denote this form by \(h\) ) is symmetric and bounded from bellow also (see Theorem 2.6 [2]}). It can be easily shown that \(h\subset \mathfrak {h},\)  but using this fact we cannot claim in general that \(\tilde{\mathcal {H}}\subset H\)  (see [2]} ). We just have an inclusion \(\tilde{\mathcal {H}}^{1/2}\subset H^{1/2}\)  (see [2]}). Note that the fact \(\tilde{\mathcal {H}}\subset H\)  follows from a condition \( \mathrm {D}_{0}(\mathfrak {h})\subset \mathrm {D}(h) \)  (see Corollary 2.4 [2]}).
However, it is proved (see proof of Theorem 4 [1]}) that relation H2 guaranties that \(\tilde{\mathcal {H}}=H.\)  Note that the last relation is very useful in applications, since in most concrete cases we can find a concrete form of the operator \(\mathcal {H}.\)

